Question

A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. Find: a) The angular frequency. s−1 b) The maximum force on the particle. N b) The maximum speed.

Answer 1

For a:v = d / Δt 
110 = 0.66 / Δt 
Δt = 0.66 / 110 
Δt = 0.006 s 
the period is: 
T = 2Δt 
T = 2*0.006 
T = 0.012 s 
the frequency is the inverse of the period. so: f = 1 / T 
f = 83.3333333 Hz (about; Hz = 1/s) 
b. T = 2π√(m/k) 
being the mass m = 200g = 0.2 kg = 2*10^-1 kg, π = 3.14 (about) and T = 0.012, k is equal to: 
0.012 = 6.28√(2*10^-1 / k) 
0.012 / 6.28 = √(2*10^-1 / k) 
0.00191082803 = √(2*10^-1 / k) 
2*10^-1/ k = 0.000003
2*10^-1 / k = 3*10^-6 
k = 2*10^-1 / 3*10^-6
k = 6.67*10^-5

now using hooke's law:
F = -kx 
F = - 6.67*10^-5* 3.3*10^-1
F = -2.20x10^-5m
F = -0.22 *10^4 N